Therefore, the line segments of 1/n and 1/(n+ 1) at the two intersections with the x axis are1/n-1(n+1).
When n= 1, 2, ..., 2004, the sum of line segment lengths is:
( 1 - 1/2)+ ( 1/2 - 1/3) + ( 1/3 - 1/4)+....+ ( 1/2004 - 1/2005) = 1- 1/2005 = 2004/2005 。
2.f( 19)=f(99)=2005, which means that the symmetry axis is: x=( 19+99)/2 = 59.
That is, -b/2a = 59, so: b=-1 18a,
F( 19)=2005, and the analytical formula is: 361a+19 b+c = 361a-19×1.
Simplification: c= 2005+ 188 1a, ——
By | c | < 1000, that is | 2005+ 188 1A |
And a is an integer, so a=-1, and substitute it into the formula 1 to get c= 124.
3. Let the quadratic function be: f(x)=ax? + bx + c，
Since f(x 1)=f(x2), that is, ax 1? + bx 1 + c= ax2？ +bx2+c, simplified as: a (x 1+x2)+b = 0-②.
Multiply the left and right sides of Equation ② by (x 1+x2) to get: a(x 1+x2)? + b(x 1+x2) =0，
So, f(x 1+x2) = a(x 1+x2)? + b(x 1+x2)+ c = 0 + c = c